In the expansion of  ${\left(7^{1/3}+{11}^{1/9}\right)}^{6561}$

General term is ${ }^{6561}{\mathrm{C}}_{\mathrm{r}}{7}^{\frac{6561-\mathrm{r}}{3}}{11}^{\frac{\mathrm{r}}{9}}$.
To make the term free of radical sign, r should be a multiple
of 9.
$\therefore \mathrm{r}=0,9,18,27,\dots ,6561$
Hence, there are 730 terms. The greatest binomial coefficients are
${ }^{6561}{\mathrm{C}}_{\frac{6561-1}{2}}{\text{ and }}^{6561}{\mathrm{C}}_{\frac{6561-3}{2}}{\text{ or }}^{6561}{\mathrm{C}}_{3280}{\text{ and }}^{6261}{\mathrm{C}}_{3279}\text{. }$
Now, 3280 are 3279 are not a multiple of 3; hence, both
terms involving greatest binomial coefficients are irrational.